Binomial expansion of $(x_{1}+x_{2}+...........+x_{k})^{n} $

Question

If we expand $$ (x_{1}+x_{2}+...........+x_{k})^{n} $$ How many terms will be there once we collect terms with equal monomials?

What is the sum of all coefficients?

I literally have no clue how to start it.

I am relating it to multinomial coefficient $$\frac{n!}{n_1!n_2!\dots n_k!}$$

but I don't know how to proceed.

Answer 1

Well, note that you need to find integers solution to $n_1+n_2+ \dots +n_k=n$, where none of them are odd.

Is this not $\binom{n+k-1}{k-1}$?

Also, if you wish to know the sum of coefficients, set $x_1=x_2=\dots=x_k=1$.

Answer 2

The sum of the coefficients will be the value when $1$ is substituted for all the $x_j$. That is $k^n$.

Using a bars and stars argument, we get $\binom{n+k-1}{n}$ terms.